HDU1312 Red and Black（DFS） 2016-07-24 13:49 64人阅读 评论(0) 收藏

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 45   Accepted Submission(s) : 34

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on
black tiles.



Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.



There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.



'.' - a black tile

'#' - a red tile

'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

————————————————————————————————————————————————————

找出@所在位置的连通块，直接DFS搜下

#include<iostream>
#include<cmath>
using namespace std;
char map[105][105];
int m, n, t;
int dir[8][2] = {{ -1, 0 },{ 0, -1 }, { 0, 1 },{1, 0 }};
 
void dfs(int si, int sj)
{
    if (si <= 0 || sj <= 0 || si > m || sj > n)
        return;
    t++;
    for (int i = 0; i < 4; i++)
    {
        if (map[si + dir[i][0]][sj + dir[i][1]] != '#')
        {
            map[si + dir[i][0]][sj + dir[i][1]] = '#';
            dfs(si + dir[i][0], sj + dir[i][1]);
        }
    }
    return;
}
int main()
{
    int si, sj;
    while (cin >> n >> m&&(m||n))
    {
        for (int i = 1; i <= m;i++)
        for (int j = 1; j <= n; j++)
            cin >> map[i][j];
        for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
        {
            if (map[i][j] == '@')
            {
                si = i;
                sj = j;
                break;
            }
        }
 
        map[si][sj] = '#';
        t = 0;
        dfs(si, sj);
        printf("%d\n", t);
    }
    return 0;
 
}